A) \[{{10}^{-1}}N/{{m}^{2}}\]
B) \[{{10}^{-2}}N/{{m}^{2}}\]
C) \[{{10}^{-3}}N/{{m}^{2}}\]
D) \[{{10}^{-4}}N/{{m}^{2}}\]
Correct Answer: B
Solution :
[b] \[\eta ={{10}^{-2}}poise\] \[v=18\,km/hr=\frac{18000}{3600}=5\,m/s\] \[l=5\,m\] Strain rate \[=\frac{v}{l}\] Coefficient of viscosity, \[\eta =\frac{\text{shearing stress}}{\text{Strain rate}}\] \[\therefore \]Shearing stress\[=\eta \times \text{strain rate}\] \[={{10}^{-2}}\times \frac{5}{5}={{10}^{-2}}N{{m}^{-2}}\]
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